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What is the difference between Best Rate of Climb and Maximum Rate of Climb?
Is there a difference between Best Rate of Climb and Maximum Rate of Climb?
From my research, best rate of climb trades ground distance for altitude (i.e. steeper climb, more altitude per unit time).  Since an aircraft cannot climb faster than its max rate, these appear to be two different labels for the same concept.
Am I missing some nuance between the two?

Generally when GA pilots talk about climb performance we speak of two different airspeed values:
Best Rate of Climb speed (Vy) gets you the most altitude per unit time (feet per minute).
When you want to get to cruise altitude quickly for maximum efficiency you'll aim for the best rate of climb so you spend the least time at lower, less efficient altitudes.
Best Angle of Climb speed (Vx) gets you the greatest altitude per unit of ground distance (feet per mile).
When you've got a FAA-Standard 50-foot-tree at the departure end of the runway you'll aim for the best angle of climb to ensure you don't wind up in the tree.
Those speeds are useful to us as pilots, but the exact rate of climb (feet-per-minute) for those speeds will vary: A fully loaded plane will climb more slowly than one that's just got the pilot and a few gallons of fuel on board, and that's where the "maximum rate of climb" enters into the discussion:
Maximum rate of climb is the number of feet per minute you can get climbing at the "best rate of climb" airspeed.
If someone is being sloppy in their usage "Maximum Rate of Climb" could mean "Best Rate of Climb" (pitch for Vy and you get what you get), but if you're being precise in your usage and really talking about the rate of climb it would mean the theoretical maximum rate of climb in feet per minute based on the current conditions and aircraft weight.

Maximum rate of climb under a given set of conditions is useful information to know if you need to clear terrain at some point on your flight path and want to be sure you can climb fast enough to do so: If you're starting from sea level and need to clear a 5000 foot mountain that's 5 minutes away but your plane can't manage more than 500 feet per minute at best-rate-of-climb speed under the current conditions you'll need to reconsider your flight plan to either avoid the mountain or climb in a circle somewhere until you can clear it.

To help remember which is which, I imagine a chart where the x-axis is distance, and the y-axis is altitude.  If distance (x) is a concern, then use Vx to minimize the x value for a given y.  If reaching a given altitude (y) quickly is your primary concern, then use Vy.  I imagine this is the origin of the notation as well.

An alternative memory device: the letter x has a lot of angles.

I think this answer would be better if best and max rate of climb were contrasted first, because that directly answers the question. Then perhaps describe best angle.

@rbp I'm not sure how you would contrast them: "Maximum" and "Best" when talking about a rate of climb are, by definition, identical: Gaining the most amount of altitude for the least amount of some other resource (usually Time, Distance, or Fuel). The real question is What rate are you trying to optimize? (Feet-per-Minute, Feet-per-Mile, or Feet-per-Gallon) - optimizing each of these rates will give a different climb profile.

Although voretaq7 has already nicely answered it, I wanted to present a picture worth thousand words.

VX
Best angle of climb speed
The greatest gain in altitude over a given horizontal distance. VX is used to clear 50' obstacles and so forth.
VY
Best rate of climb speed
The greatest gain in altitude over a given amount of time. VY is used on normal takeoffs and such.

I always liked this image. What's the difference between Vx and Vy?  "At Vy you'll crash through the middle of the tower, but at Vx you'll just clip the antenna masts!"

user2168, I don't like this picture! Why the second aircraft is at the same height? Where is the "greatest altitude gain"?

Depends what is best in the particular case. Generally, best means highest climb speed, but there might be other things which should be optimized:

highest flight path angle: This might be desired to avoid noise on the ground, or to escape unfriendly fire near the airport. For propeller aircraft, the optimum lift coefficient is
$$c_L = \frac{T\cdot\pi\cdot AR\cdot \epsilon}{4\cdot m\cdot g} + \sqrt{\left( \frac{T\cdot\pi\cdot AR\cdot \epsilon}{4\cdot m\cdot g}\right)^2 + \pi \cdot AR \cdot \epsilon \cdot c_{D0}}$$
lowest fuel consumption per altitude gained: The highest climb speed is reached with full power, but maybe a lower power setting is more economical. In general, this is very close to the schedule with maximum climb speed. For propeller aircraft, the optimum lift coefficient for highest climb speed is the same as for minimum energy loss:
$$c_L = \sqrt{3 \cdot \pi \cdot AR \cdot \epsilon \cdot c_{D0}}$$
highest energy gain (the sum of both altitude and speed gain is optimized): This is desired when you want to reach a point far up in a hurry, like a supersonic interceptor would.

The graph below shows lines of equal total energy (black, dashed) and of maximum altitude for given climb speeds over true airspeed (blue, solid). The red line connecting the tops of the blue lines gives the flight schedule for fastest altitude gain, and the green line cutting through the blue lines at their maximum of total energy gives the schedule for the total energy gain climb. The higher you fly, the wider is the difference in optimum speed between both.

Nomenclature:
    $c_L \:\:\:$ lift coefficient
    $T \:\:\:\:$  thrust
    $m \:\:\:\:$  aircraft mass
    $g \:\:\:\:\:$  gravity
    $\pi \:\:\:\:\:$  3.14159$\dots$
    $AR \:\:$  aspect ratio of the wing
    $\epsilon \:\:\:\:\:$  the wing's Oswald factor
    $c_{D0} \:$  zero-lift drag coefficient

Peter Kampf: single-handedly proving math is used after high school!

Kinda overkill though, maybe?  (The question was really about semantics, not math...)

@MichaelHall Call it semantics, I call it physics. There are several "best" climbs, depending on what needs to be optimised. And using math makes the answer much clearer.

The question was whether two different adjectives, (best and maximum) when applied to a single noun, (rate) could be used interchangeably and mean the same thing in the context of aviation.  You are welcome to call that physics, but that pretty much meets the definition of semantics.  Look, we all know you are brilliant, but good teachers teach to the level of their students.  Math may make some answers more clear, but that isn't universally true when it comes to simple questions.

@MichaelHall I give up. You seem not to understand "nuance".

I understand nuances, but from what I perceive as the OP's level of aviation knowledge I still think your answer was overkill.  JMHO.

No.
Best and Maximum (Rate of Climb = ROC) are usually$^1$ used as synonyms.
For completeness only: There is also the AOC (Angle of Climb), which is just another quantity to describe your flightpath (the angle of your climb).
In order to reach the maximum of:

ROC you have to maintain the airspeed $V_y$
AOC you have to maintain the airspeed $V_x$

Here is a nice picture illustrating the two
How to remember the two: x is before y in the alphabet. On take-off (for small planes at least), $V_x$ is used first because you want to climb as steep as possible to be free of any obstacles. Once you are free, you change to $V_y$ in order to climb faster.
1: I say "usually", because "best" may refer to something else, like "best in terms of fuel consumption", "best in terms of flying directly into the tree in front of us" etc. but then the meaning is unambigious

Is there a difference between Best Rate of Climb and Maximum Rate of Climb?

There's no difference, end of story.  This question inspired some good answers but many of them were really better suited to some other question!

From my research, best rate of climb trades ground distance for altitude (i.e. steeper climb, more altitude per unit time).

This is incorrect, the best or maximum or highest climb rate does not occur at the same airspeed as the steepest climb.  The latter always occurs at a lower airspeed.  Maximizing the vertical distance travelled in relation to the horizontal distance travelled gives you the steepest climb, not the maximum rate of climb.

VY is similar to a transport aircraft v2 which will also get you to 1500 feet above the ground in the least amount of time possible ,VY will get you up to a pre-determined altitude usually  between 1000 to 1500 AGL. The objective of getting to 1000 to 1500 feet as soon as possible is at that altitude you have more altitude to maneuver if an emergency occurs (example an engine failure)
after obtaining such an altitude VY should no longer be maintained and aircraft should be transitioned to a cruise climb to increase visibility and engine cooling usually about 500 feet a minute, I hope this helps anyone that was looking for a clear explanation ....Good luck and stay safe thanks Joe

A jet aircraft flying at V2 with all engines operating is way too slow.  It is more closely related to Vxse.  V2 and Vxse are not the same speeds however.

Despite the simple explanation that differentiates between the two types of climb (Vx and Vy) based on shortest distance and shortest time does not give clear  assimilation unless you consider the concept of total energy and drag. At same power setting, thrust or total energy remains the same during both situations but the drag is more during best angle of climb which creates loss of energy and thereby lesser distance is covered as well as lesser height is gained compared to that of flying with best rate of climb speed. So in order to cross above an obstacle after take-off, we are intentionally trading off some height gain and distance covered to take the aircraft to a point above the obstacle. But, had we climbed at best rate of climb after take off, we would gain more height and distance within same period of time but ultimately hit the obstacle.


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Do pilots use the pedals in flight on planes with a sidestick (Airbus)?
Do the pilots flying sidestick aircraft with fly-by-wire control systems, such as the Airbus A320, use the rudder pedals while in the air? Or does sidestick input affect both the rudder and the ailerons?

The rudder pedals are not usually used in flight, even with the autopilot off.
The rudder on the Airbus A320 can be controlled via a mechanical connection to the rudder pedals in the flight deck, but also electrically via the fly-by-wire mechanism. The following graphic shows an overview of the rudder control:

(Airbus A320 FCOM - Flight Controls - Description)
As you can see the sidestick input goes to the ELAC (ELevator Aileron Computer) and from there a Yaw Order is sent to the FAC (Flight Augmentation Computer). This will then be passed on to the rudder via the Yaw Damper Servo Actuator.

Electrical Rudder Control
The yaw damping and turn coordination functions are automatic. The
ELACs compute yaw orders for coordinating turns and damping yaw
oscillations, and transmit them to the FACs.
Mechanical Rudder Control
The pilots can use conventional rudder pedals to control the rudder.

(Airbus A320 FCOM - Flight Controls - Description)
Since the yaw damping and turn coordination are fully automatic, there is usually no need for the pilot to use the rudder pedals in flight.
There are however some situations, where manual rudder input via the pedals is required:

Takeoff Roll: The rudder pedals move both the rudder and the nose wheel steering (NWS, see "NWS ORDER" in graphic above) on the ground to maintain directional control during takeoff and landing. After initiating the takeoff roll, the pilot flying (PF) will use the rudder pedals to control the aircraft:

The PF should use pedals to keep the aircraft straight. The nosewheel steering authority
decreases at a pre-determined rate as the groundspeed increases (no more efficiency at
130 kt) and the rudder becomes more effective.

(Airbus A320 FCTM - Normal Operations - Takeoff)

Landing: When landing the aircraft in a crosswind the direction in which the aircraft points is not aligned with the runway. To align the aircraft before touchdown (called de-crab), the PF needs to apply rudder input via the pedals during the flare:

During the flare, rudder should be applied as required to align the aircraft with the
runway heading. Any tendency to drift downwind should be counteracted by an
appropriate lateral (roll) input on the sidestick.

(Airbus A320 FCTM - Normal Operations - Landing)
This is the only time (for normal operations) where the rudder pedals are actually used in flight.

Engine Failure: After an engine failure the asymmetric thrust of the remaining engine needs to be overcome with the rudder. For an engine failure during the takeoff (but after V1), this needs to be done manually with the rudder pedals:

On the ground: Rudder is used conventionally to maintain the aircraft on the runway centreline. [...]
When safely airborne: [...] Since
the lateral normal law does not command the full needed rudder surface deflection,
the pilot will have to adjust conventionally the rudder pedals to center the beta
target. [...]

(Airbus A320 FCTM - Abnormal Operations - Operating Techniques)
Once the aircraft has accelerated, the rudder is trimmed and no pedal input is required any more.

"For an engine failure during the takeoff (but after V1), this needs to be done manually with the rudder pedals" Would this not also be required in the event of an engine failure before V1 in order to maintain the centerline until the remaining engine can spool down and/or reverse thrust deployed? For example, this video shows an A330 engine failure before V1 and you can almost immediately see hard rudder deflection to counter the asymmetric thrust and push the nose back toward the centerline.

@reirab Sure, the manual rudder input is also required before V1, but as you say, once the other engine has spooled down this would fall under point 1 (Takeoff Roll). I only specifically mentioned the case after V1 because you would still need manual pedal input in the air, which is what OP was asking about.

Side stick or no, once airborne, you never touch the rudder pedals in any airliner when flying except during landing where independent rudder inputs are used to keep the airplane lined up during the flare.  The yaw damper takes care of rudder coordination for normal flying; that is, in addition to controlling yaw excursions, it compensates for adverse yaw for you.  The other main use of the pedals is for asymmetric thrust conditions from engine failures.
You use rudder on takeoff for nosewheel steering control, and you keep your feet on the pedals during the initial part of the departure in case an engine quits, but you never actually make any inputs while flying outside of landing, and I used to just put my feet on the floor after the departure to avoid applying pedal without realizing it during hand maneuvering, putting my feet back on the pedals on final.

Sidestick input does NOT control both the yaw and roll axes. Pilots of sidestick aircraft still use the rudder pedals to control yaw and they use the sidestick for roll. This allows them to perform crosswind landings and sideslips in addition to coordinated turns.

As far as coordinated turns, you are right in the case of side-stick aircraft that do not employ a fly-by-wire system like a Cirrus. John K and Bianfable have pointed out that the computer does a lot more work for you in newer Airliners.

I wouldn't necessarily say "newer". I don't think you'll find any large swept wing aircraft without yaw dampers, dutch roll issues being inherent to the configuration, so even in a jet from the 60s you wouldn't be using rudder to fly and let the YD do it.  In the very early days of autopilots, they were 3 channel with the AP running rudder as well (like the B-17).  Jets went with standalone YDs because that active yaw control function is needed all the time not just when AP is engaged.

@JohnK -- about the only jet I can think of that wouldn't have been that way back then were the early KC-135s, because they had no YD to start with

Also no lift dumpers!  There was a PBS documentary called Boomer And The Crew Dogs about a KC135 squadron in Plattsburg NY https://video.pbswisconsin.org/video/boomer-and-the-crew-dogs-1992-za7afy/  and I remember a great clip of a KC doing touch and goes, with shots zoomed on the gear, and they would do gentle hops and skips once or twice on some landings, there being no landing spoilers.  I wonder if the 707/KC had a sufficiently mild dutch roll mode to get away with no YD.  On a lot of jets, perhaps most, it's quite dangerous.

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How can I use electric motor data with my prop data to obtain a power available versus velocity curve?
I'm not entirely sure this is appropriate for this forum but it's worth a shot...
I have propeller data ($C_T$ , $C_P$ , J , and $\eta$) and very very limited electric brushless motor data (motor diameter, shaft diameter, Kv, voltage range, max amperage) I am trying to put together a power available curve (P vs V) but I'm not exactly sure how to use the data I have to do so.  How do I get shaft power from my engine data?  Or what can I use?

So your question is about the shaft power vs the voltage? I mean the propeller data have nothing to do here, correct? If so, here is some information, but I'm not sure this is complete.

No.  Power available to the aircraft verses velocity.

For electric motors, torque is a function of amperage and rotational speed is a function of voltage.

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Causes of yaw in slip/skid turn
Why does the plane yaw when the rate of turn is not suitable to bank angle in a turn? In a slipping/skidding turn, there is an imbalance of centripetal/ centrifugal force. How does that cause a yaw?

Actually-- should have asked first-- what do you mean by "yaw" anyway?  Are you talking about a change in yaw rate, to some inappropriate value?  Are you talking about a deflection of the yaw string?   Virtually every turn involves yaw, it's not only a feature of slipping or skidding turns.  So that makes your question a bit hard to understand.

This just has to be a dupe…. Have you looked for answers?!

Yaw : nose turning into turn in a skidding turn, putt of turn in slip turn

It appears to me that this question has essentially been replaced by a similar but potentially better thought-out question https://aviation.stackexchange.com/questions/97114/what-causes-a-slip-skid-turn .  The present question should probably be closed or deleted.  Normally we discourage deletions of questions with existing answers, but in this case it might be appropriate.

@quietflyer, I agree these are basically identical questions, so why did you put so much effort into answering both of them?  We ought to be voting to close duplicates, not assist in their propagation.  And really both new questions have been answered to death already.  Voting to close.

Does this answer your question? What is a skidding turn (vs slipping turn)?

@guykatz -- not terribly important, just a bit of housekeeping, but your other question was better and since I answered it too, maybe you should "unaccept" my answer to this one, and that would allow me to delete my answer, which I'd like to do.  Then if you wish you might want to delete this question, since you have essentially replaced it with a better one, and that way you would wouldn't be deleting a question with an existing answer, which I think would give you negative hit on reputation.  None of this matters much but I'd prefer to see your other question --(ctd)

(ctd) eventually re-opened, and this might help.

Why does the plane yaw when the rate of turn is not suitable to bank
angle in a turn?

If you mean why does the plane slip or skid when the rate of turn is not suitable to bank angle in a turn, then see item #5 below.

In a slipping/skidding turn, there is an imbalance of centripetal/
centrifugal force. How does that cause a yaw?

It really doesn't.
Here's a better way to think about slip and skids and the resulting imbalance in forces--
The "causal chain" goes like this--

Airplane nose is not fully aligned, yaw-wise, with actual instantaneous direction of travel, generally due to some aerodynamic yaw torque (for example due to "adverse yaw" from the deflected ailerons and/or from a non-zero roll rate, or due to an intentional rudder input by the pilot) that counteracts the aircraft's natural "weathervane stability" or "directional stability".  Yaw rotational inertia can also play a contributing role as a turn is being initiated (or exited), but this effect is generally slight.

(This "misalignment" between nose and instantaneous direction of flight path is sort of but not exactly the same as the yaw rate being "wrong" for the bank angle.)

Side of fuselage is exposed to free-stream relative wind

Airflow hits side of fuselage and generates a real, tangible aerodynamic force

The net aerodynamic force vector is no longer pointing straight "up" in the reference frame of the banked aircraft, as it would be if the wings' lift vector were the only unbalanced aerodynamic force at play.

This causes a reduction (or increase) in turn rate so the turn rate is now "too low" (or "too high") for the bank angle.

This also causes the slip-skid ball to ride off-center.

Note that either item 4 or item 5 can be viewed as the fundamental "cause" of item 6, depending what reference frame we use.  Item 4 rather than item 5 is best viewed as the cause of item 6 if we are using a valid inertial reference frame -- e.g. the earth, not the moving airplane-- in which case our explanation should not invoke the fictitious "centrifugal force".
Note that for the exact same degree of "misalignment" between the aircraft nose and the actual instantaneous direction of the flight path-- i.e. the exact same displacement of a "yaw string"-- the resulting amount of aerodynamic sideforce generated by the airflow against the fuselage, and the resulting displacement of the slip-skid ball, and the resulting change in the turn rate, will vary according to the shape of the fuselage and the amount of side area therefore presented to the airflow.  So a given deflection of the yaw string will cause a different deflection of the slip-skid ball in various different aircraft.
Note that by putting item #1 at the very front of the "causal chain", we're well equipped to understand why appropriate rudder inputs are the key to preventing slips and skids.
Note also that this perspective works well even for extreme cases, even including 90-degree-banked "knife edge" linear flight such as we see at airshows.  Some other "explanations" that we read in pilot training materials tend to break down in such extreme cases.

There are many related ASE answers that go into this topic, especially the exact cause of the deflection of the slip-skid ball-- will post some more when I get a change.

There's another perspective that can be taken here -- at least in a steady-state situation, the only way the yaw rate (or the turn rate) can be "wrong" for the bank angle is if something is creating a real aerodynamic force that speeds or slows the turn rate.  And what would that "something" be?  Item #4 above.  So yes it's "true" that we get a slip or skid if yaw rate or turn rate is "wrong" for the bank angle-- but as for why-- see this answer.  Will try to condense this comment into a useful footnote--

"There are many related ASE answers that go into this topic, especially the exact cause of the deflection of the slip-skid ball".  Recommend you find one and VTC.


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What causes a slip/skid turn?
Is the rate of turn being too low/high for a given bank angle the cause of slip/skid turn? Or is it yawing too much inside/outside turn causing an imbalance in centripetal/fugal force and thus the slip/skid turn?
According to the FAA handbook: “In a slipping turn, the aircraft is not turning at the rate appropriate to the bank being used, since the aircraft is yawed toward the outside of the turning flight path. The aircraft is banked too much for the ROT, so the horizontal lift component is greater than the centrifugal force.”
Is that statement true?

Or this one:  https://aviation.stackexchange.com/questions/77528/what-is-missing-from-these-diagrams-of-the-forces-in-slips-and-skids?noredirect=1&lq=1

Or this one:  https://aviation.stackexchange.com/questions/79338/what-are-the-forces-present-in-a-coordinated-turn?rq=1

(Hint, if you are on a full screen PC and not your phone, look at "related" questions on the right side.  Or search by key words...)

One question cited as duplicate is closed, other is not really a duplicate.  Should be reopened.

@quietflyer prefer to let it rest.  Imagine a new pilot turning uncoordinated and not checking airspeed. Double drag.  Not good.

Yes, please, just let it rest.  There is a plethora of information here and on other sites for those unfamiliar.  Nothing new and productive can some from hammering on it more...

Or this one:  https://aviation.stackexchange.com/questions/75380/what-exactly-is-a-coordinated-turn?rq=1

@quietflyer  Seriously, between your own lengthy answer and ELEVEN comments (mostly to self) in the question linked immediately above, what more can you possible have to say?!  And in how many more questions on this site could one glean an understanding of the basic subject?  (I found four immediately in the related questions links on the right side.)  There is plenty of information out there for anyone to find with just a little effort, the OP isn't asking anything fresh or new here.

Is the rate of turn being too low/high for a given bank angle the
cause of slip/skid turn?

One could say that the rate of turn being "too low" or "too high" for a given bank angle is a symptom of something else, which is really the more fundamental cause of slip or skid in a turn.

Or is it yawing too much inside/outside turn causing an imbalance in
centripetal/fugal force and thus the slip/skid turn?

Yes, if we understand the word "yawing" in this context to mean not the actual direction and rate of yaw rotation, but rather the condition of flying in such a manner that a yaw string would stream toward the inside or outside of the turn.  I.e., the condition of flying with the nose pointing toward the inside or the outside of the aircraft's actual instantaneous direction of travel.

According to the FAA handbook: “In a slipping turn, the aircraft is
not turning at the rate appropriate to the bank being used, since the
aircraft is yawed toward the outside of the turning flight path.

I'm actually surprised that the FAA published such an enlightened statement.  In the past, in flight training materials, including materials from the FAA, the emphasis is entirely on the turn rate being "wrong" for the bank angle.  But the statement quoted above seems to recognize that the fundamental cause of this "mismatch" between the turn rate and the bank angle is the fact that the aircraft is being flown in a "yawed" condition (in the sense described above), i.e. the nose is not aligned with the actual instantaneous direction of the flight path, so the airflow is striking one side of the fuselage, generating a real aerodynamic force that causes an alteration in the the turn rate.

The aircraft is banked too much for the ROT, so the horizontal lift
component is greater than the centrifugal force.”
Is that statement true?

This is where things get really confusing.  That final quoted statement is technically true, but very misleading.  We'll revisit that at the end of this answer.
Ultimately I'd suggest that the "causal chain" goes like this--

Airplane nose is not fully aligned, yaw-wise, with actual instantaneous direction of travel, generally due to some aerodynamic yaw torque (for example due to "adverse yaw" from the deflected ailerons and/or from a non-zero roll rate, or due to an intentional rudder input by the pilot) that counteracts the aircraft's natural "weathervane stability" or "directional stability".  Yaw rotational inertia can also play a contributing role as a turn is being initiated (or exited), but this effect is generally slight.

(This "misalignment" between nose and instantaneous direction of flight path is sort of but not exactly the same as the yaw rate or turn rate being "wrong" for the bank angle.)

Side of fuselage is exposed to free-stream relative wind

Airflow hits side of fuselage and generates a real, tangible aerodynamic force

The net aerodynamic force vector is no longer pointing straight "up" in the reference frame of the banked aircraft, as it would be if the wings' lift vector were the only unbalanced aerodynamic force at play.

This causes a reduction (or increase) in turn rate so the turn rate (and likewise the yaw rate) is now "too low" (or "too high") for the bank angle.

This also causes the slip-skid ball to ride off-center.

Note that either item 4 or item 5 can be viewed as the fundamental "cause" of item 6, depending what reference frame we use.  Item 4 rather than item 5 is best viewed as the cause of item 6 if we are using a valid inertial reference frame -- e.g. the earth, not the moving airplane-- in which case our explanation should not invoke the fictitious "centrifugal force".
Note that for the exact same degree of "misalignment" between the aircraft nose and the actual instantaneous direction of the flight path-- i.e. the exact same displacement of a "yaw string"-- the resulting amount of aerodynamic sideforce generated by the airflow against the fuselage, and the resulting displacement of the slip-skid ball, and the resulting change in the turn rate, will vary according to the shape of the fuselage and the amount of side area therefore presented to the airflow.  So a given deflection of the yaw string will cause a different deflection of the slip-skid ball in various different aircraft.
Note that by putting item #1 at the very front of the "causal chain", we're well equipped to understand why appropriate rudder inputs are the key to preventing slips and skids.
Note also that this perspective works well even for extreme cases, even including 90-degree-banked "knife edge" linear flight such as we see at airshows.  Some other "explanations" that we read in pilot training materials tend to break down in such extreme cases.
Now, returning to --

The aircraft is banked too much for the ROT, so the horizontal lift
component is greater than the centrifugal force.”

This statement is technically true, but unhelpful.  The centrifugal force is nothing more and nothing less than the exact mirror-image of the actual centripetal force, and the centripetal force is simply the horizontal component of the net aerodynamic force generated by the aircraft.
So to say that the "horizontal lift component is greater than the centrifugal force", is just a really complicated and obtuse way of saying that the aircraft is generating some other aerodynamic horizontal force component that is opposing the horizontal force from the lift vector, and also is slowing the turn rate.  That other horizontal force component is the airflow striking the side of the fuselage, on the side that is inside the turn.  So the specific situation being described here is a slip.  And the point that is left out of the quoted description, is the simple fact that the net aerodynamic force vector is no longer pointing "straight up" in the aircraft's own reference frame, as it would be if the lift vector from the wing, and gravity, were the only forces at play.
NOTE that this aerodynamic sideforce from the airflow striking the side of the fuselage is missing from all the diagrams depicting slips and skids in this related ASE question, even though the resulting "imbalance" between centrifugal force and the "horizontal component of lift" is depicted.  This is a fundamentally flawed way of illustrating the forces at play in slips and skids.
(Link: What is missing from these diagrams of the forces in slips and skids? )

Thank u very much for the answer. It cleared up the confusion for me. The FAA handbook really confused me with the whole :” ROT not suitable for the bank angle “ statement . Thx again

Can the turn radius stay the same as the forces reach equilibrium - the centripetal and centrifugal, but the lift vector is oriented incorrectly for the rate of turn? Imagining an aircraft on a string - in a slip, if the aircraft could rotate into the turn it would be coordinated? (almost as if it were being pulled around the circle by its tail?)

@RyanBW  -- your question is not really making sense to me.  As I see it, if the aircraft "rotates into a turn" during a slip, the slip stops (and in extreme cases may become a skid).  To me the fundamental issue in discerning a slip or a skid or neither, is simply asking whether the nose is pointing toward the outside of the instantaneous direction of the flight path, or the inside, or is fully alligned with the instantaneous direction of the flight path.  So if you change the direction the nose is pointing, you change the extent of the slip or skid.

@RyanBW -- so in other words, there's no such thing as slipping turn, with the nose somehow rotated to point toward the inside of a turn.  Does that help?  Or maybe I am somehow missing something in your question.  I hope my answer helped, I think it is one of my better ones, (better than some my answers to questions that this is claimed to be a duplicated of) -- not too ungodly long but hitting the important points thoroughly -- if you have any specific questions about it I would be happy to try to answer them.  If we end up going to chat that's fine too.

The pilots answer might be: use your rudder to fix it. Step on the ball.
If you are uncomfortable with your bank angle, you can also roll out of it a bit.  Ailerons and rudder work together here.
But the theory may be more difficult to wrap your head around, but I will try:
The wing is "boss" when it comes to generating flight forces.  Lift is the most powerful force your plane creates.  Thrust is second. (And gravity is there too).

"the horizontal component is greater than the centrifugal force"

The horizontal lift component creates the centripetal force.  The opposite "centrifugal force" is the result of the plane being pulled (accelerated) to the side.  If the plane is banked too steeply and not yawing enough, it is slipping.  The ball is to the inside.  If the plane is yawing too much and not banked enough, it is "skidding" and the ball is to the outside.  You want the correct yawing and banking combination to center "centrifugal force" right through the floor$^1$, which puts the ball right between the marks on the curved tube inclinometer.
Adverse yaw, from the outside aileron being deflected down, tends to pull the nose outside the turn.  Now you are in a "slipping" turn.  Ball is to the inside.  "Coordinate" by stepping on the ball.
The tail of a properly designed airplane should "weathervane" the nose into the wind, and the rudder helps, if needed.
Some aircraft have more adverse yaw tendency than others.  But centering the ball in turns with the rudder is very good because it reduces drag and is most comfortable for passengers.
$^1$ the gravity component also affects the position of the "ball".  It is the resultant vector of gravity and centrifugal force that centers the ball.  This is what you feel "in the seat of your pants".


text:
How does conical camber decrease roll due to sideslip?
I was reading Ray Whitford's Design for Air combat and came upon a passage that said that on the SR-71 conical camber "gives a useful reduction in the otherwise very high rolling moment due to sideslip inherent in the highly swept delta wing."
I could not understand how this would happen and dug a little deeper to find a paper  NACA RM A56DO2 where as per my understanding the opposite is said.
If what the book says is true, then an aircraft with a conical camber wing flying with the relative wind from the left (sideslip) should roll to the right with less torque/ rate of roll compared to a normal delta wing.
Here's what the paper says:
x-axis= Cl= rolling moment coefficient
y-axis= Angle of attack
Here the wing is facing relative wind 6° off to the left. We can see that the conical cambered wing with relative wind from the left produces a stronger roll to the right compared to a normal wing. The book says the opposite.
The paper further adds that the conical camber increases maximum effective dihedral. Which will again promotes stability. So am I having a brain fart or is this a contradiction?
P.S Here's the entire graph page:

Thanks for showing the whole graph page, that helps.

I'd be interested in seeing a new question that simply asked,  "Does conical camber in fact decrease rolling moment due to sideslip at positive angles-of-attack, and if so, how does this happen, aerodynamically speaking?"   Maybe will have to ask it--

I edited the opening sentence because your quote did not exactly match the quoted sentence from the book.  If you like your wording better, you can put it back, but I'd suggest removing the quotes.

I created a chat room to discuss further-- feel free to post there--https://chat.stackexchange.com/rooms/145487/a-room-to-talk-about-how-conical-camber-affects-roll-torque-generated-by-sideslip

... using my desert cooler as a wind tunnel

That's the spirit!
Let's first evaluate the degrees sweep of the B-58 Hustler and F106 Delta Dart vs the NACA paper.  The former are 60 degrees sweep and the paper covers 45 degree sweep.
This becomes important because highly swept delta wings use vortex lift, which can be easily broken in a sideslip.  This may explain why the book and the paper provided conflicting results.  Add in the effects of engine nacelles acting as fences and you have a fairly complicated situation even before Mach effects.
From personal paper airplane experience, 45 degrees seem to be in a "transition zone" between straight wing and slender (highly swept) wing cross wind behavior.
Straight wings with dihedral effect tend to roll away from a cross wind, while slender deltas roll into due to loss of upwind wing lift.
Flying a model Delta Ray, which did fine > half throttle, a real "bricking" tendency was noted banking low and slow (you know how).  The feeling at the RC forum was the wingtips may have been better pointed downwards!

I created a chat room to discuss further-- https://chat.stackexchange.com/rooms/145487/a-room-to-talk-about-how-conical-camber-affects-roll-torque-generated-by-sideslip

I was reading Ray Whitford's Design for Air combat and came upon a
passage that said that on the SR-71 conical camber "gives a useful reduction in the otherwise very high rolling moment due to sideslip inherent in the highly swept delta wing."

...

If what the book says is true, then an aircraft with a conical camber
wing flying with the relative wind from the left (sideslip) should
roll to the left or at least the rate of roll to the right should be
reduced.

(Bolding added)
There's a huge difference between simply reducing the rolling moment due to sideslip, and reversing the rolling moment due to sideslip!  Delta wings are well known to have a strong positive rolling moment due to sideslip, and in the absence of a clear and direct statement, it would be unreasonable to assume that the book was trying to convey that the direction of this rolling moment would be reversed by introducing conical camber, as if the wing had a great deal of anhedral.  Nothing that you've quoted supports that conclusion.  The bolded sentence in particular clearly references a reduction in rolling moment due to sideslip, not a reversal in the direction of the rolling moment due to sideslip.1
However, you are correct in noting that there is a contradiction between the idea expressed in the book-- namely that introducing conical camber on a delta wing reduces the rolling moment due to sideslip-- and the graph from the NACA paper, which shows that-- at least between about 4 and 10 degrees of positive angle-of-attack-- delta wings with conical camber show a stronger rolling moment due to sideslip than does a delta wing without conical camber.
Footnotes:

It would not be unreasonable to imagine that a design feature that reduced the rolling moment due to sideslip over most of the positive angle-of-attack range, might reverse the direction of the rolling moment due to sideslip at very low positive angles-of-attack (i.e. very near zero).  For example, that's what we'd expect to see if we added a mild amount of anhedral to a delta wing.  So we need to be a bit careful about blanket statements as to whether a given design feature reduces the rolling moment due to sideslip, or reverses it-- it may well depend upon what part of the flight envelope we are looking at.  This is particularly relevant in the case of a very-high-speed aircraft that will cruise at very low angles-of-attack.  However that nuance is clearly not the root of the observed discrepancy between the two sources.   Unlike what we'd expect to see in the case of a delta wing with anhedral, on the plot you've provided from the NACA paper, it appears that the delta wings with and without conical camber all experience nearly zero rolling moment due to sideslip at angles-of-attack near zero degrees.

Obviously we're speaking in broad generalities here.  The effect (on rolling moment due to sideslip) of adding conical camber to delta wing could well depend on the aspect ratio of the wing.  Look at photos of hang gliders, especially the old "Rogallo" designs with tremendous amounts of conical camber.  This can be viewed as essentially adding dihedral to the inboard portion of the wing, and adding anhedral to the outboard portion.

(ctd) Highly tapered (low-aspect) ratio designs have so little area in the outboard (anhedral) portions that resulting shift in "effective dihedral" is distinctly in the positive direction; with less tapered (higher-aspect) ratio designs the reverse appears to be true, so that the effective dihedral is shifted in the negative direction and reaches negative values at low (but still positive) angles-of-attack.

(ctd) We don't see anything like this in the NACA graph, and I'm not sure the geometry of the conical camber in the Rogallo wing is really very similar to geometry of the conical camber that NACA is graphing. In the Rogallo case, the leading edge remains linear, while the camber is created by sail billow such that the trailing edge rises from root toward mid-span area and then descends again back toward the tip. That's not the geometry we see when we look at a SR-71!

The second quoted passage from the question was changed during the time that I was creating and posting this answer-- the question no longer interprets the passages in the book to suggest that conical camber creates a reversal in the direction of the roll torque due to sideslip.

Nonetheless I'm just going to let the answer stand as is, at least for now.

@Battery -- I can think of a reason why adding conical camber to the leading-edge geometry might decrease the rolling moment generated by sideslip.  Think of what the wing looks like in an oblique view (partly head-on, partly from the side).  Think of how it therefore "looks" to a sideways component in the airflow.  Doesn't address why the NACA paper apparently came to the opposite conclusion though.  Not quite ready to add this content to an edit to answer yet, but stay tuned.

Thank you for all the help on the post! I've been thinking, the NACA paper and the book maybe talking about sideslip with different origins. The book maybe talking about crosswind conditions when you sideslip by slightly rolling into the wind (changes AoA on wings). While the paper is talking about pure sideslip like when there's an unexpected change of wind direction or a gust. I'm also trying to write to the author, meanwhile I'm to trying to build a small model aircraft with conical camber wings and using my desert cooler as a wind tunnel.

I don't think the origin of the sideslip should matter.  Sometime maybe I'll find time to take some pictures of a simple model to try to explain why I think adding conical camber to the leading-edge geometry might make an effect rather like adding some anhedral to the wings.  Feel free to "ping" me ("@username", or just leave another comment here) to let me know what you find with further research, we could always set up a chat room to discuss further--

I created a chat room to discuss further-- https://chat.stackexchange.com/rooms/145487/a-room-to-talk-about-how-conical-camber-affects-roll-torque-generated-by-sideslip


text:
Is it possible to use a plane's entire wing as an elevon?
Would it be feasible to build an RC plane using the entire wing as an elevon, with each wing independently moved up and down with servos? For roll control, they would be moved in opposite directions; for pitch, they would move together. (Of course, in reality, both will be done at once.)
I'm asking here rather than on the drone site because the question is mostly about theory, and this site seems a lot more active than the drone one.

(Obviously not to scale)
Green: Propeller
Red: Movable wings
Yellow: Center of gravity
Blue: Fixed wing

You mean aileron? If the entire wing is moving differentially, the pitch trimming forces have to come from somewhere else.

@JohnK I was thinking of moving both wings together for pitch. Would that work?

If you implement it on a conventional airplane where CG is in front of the wing, pitching the wing up will cause a nose down, which conflicts with the pitch-up of the wing. To make it work, the pivoting wing must be in (near) front of the CG and there needs to be another pair of wings (of similar size) further after the CG to stabilize the pitch.

@user3528438 something like the diagram I added?

@user3528438 -- I don't think it matters much to the effectiveness of the proposed pitch control system, whether the aircraft is set up so that the tail needs to produce an upforce (CG behind wing's "center of lift") or a downforce (CG ahead of wing's "center of lift") in normal flight.  At most, with the latter, you might have a transitory small adverse pitch  torque when initiating a pitch change, that might basically just make it seem as if the aircraft had more pitch rotational inertia than was actually the case.

But meanwhile the flight path is curving and that soon creates the flow direction at the tail needed to create the needed pitch torque to initiate the required pitch rotation.  Well, at any rate it's a fairly complicated and interesting subject-- but it works out in actual practice--

So from your picture it seems like you are designing a propeller version of an Eurofighter  Do you have also a vertical tailplane?

@sophit it's not quite like the Eurofighter, because it's not a delta wing. Will I need a vertical tailplane ?

I drew the tail wing too big. The movable wings will be the main ones.

Just joking. I suppose you need a vertical tailplane as well for stability

Sorry to post this a little late, but have just now found that this question appears to be a duplicate of https://aviation.stackexchange.com/questions/46195/could-flight-control-be-achieved-by-rotating-a-wing-entirely

Would it be feasible to build an RC plane using the entire wing as an
elevon, with each wing independently moved up and down with servos?

Yes, such designs exist.  They are most often seen in radio-controlled slope gliders.  The key is that there is also a horizontal tail which is fixed.  Otherwise there would be no pitch control-- moving the wings together in the same direction would accomplish nothing in a tailless design.1
If the wings move in the opposite direction for roll control, but pitch control is accomplished with elevators on the horizontal tail (or with an all-moving horizontal tail), then the design is described as a "wingeron" design.  (The term is a "portmanteau" of "wing" and "aileron".)  With proper gearing, "wingeron" designs can be built with only one servo driving the wing movement, or two servos can be used.
If the wings move in opposite directions for roll control and also move together for pitch control, as posited by the O.P. here, then the aircraft is described as a "pitcheron" design.  This word is "portmanteau" of "pitch" and "wingeron".  "Pitcheron" designs always need two servos to control the wings.
Here is a photo of a "Pica" radio-controlled slope-soaring glider-- a "pitcheron" design, with no elevators or other moving parts on the horizontal tail (actually a V-tail in this particular case).  (Source: "Slopeflyer" website )

Here is a photo of a Sig "Samurai" radio-controlled slope-soaring glider-- a "wingeron" design, with elevators on the horizontal tail (again, actually a V-tail in this case).  (Source: "Wanderings" website.)

Note the absence of ailerons in all the photos.
Here's a link to a YouTube video of a Sig Samurai "wingeron" glider in action.  The internal workings are briefly visible near the start of the video.  One single servo controls both wings for roll control.  One more servo to control the elevator function of the tail would have been sufficient.  In this particular case the glider is set up with a second servo for the tail (making three servos in all) to allow the moving surfaces to act as "ruddervators" (as in a Beechcraft Bonanza) rather than just as elevators, but the rudder function is really not necessary in a fast aircraft of this nature, and certainly is not a significant source of roll control, since the wing has neither dihedral nor sweep to create slip-roll coupling.  As noted in the video, this particular design is often built with just two servos, omitting the rudder function of the tail surfaces.
Link to a related question on a model airplane discussion forum: Pitcheron & Wingeron. What's the difference?
Note than in the case of a powered aircraft, a "pitcheron" design will create a "thrust vectoring" effect that might be less than optimal, depending on the specific details of the design.  When the pilot increases the angle-of-attack of the wing by moving the control stick aft, he is essentially tilting the thrust vector slightly downwards relative to the wing, which seems less than ideal.  However, if the thrust line can be configured to pass through or near the CG of the aircraft regardless of wing position, then this is not the same as an increase in "downthrust" angle, which would generally have much a more unfavorable effect on handling.  Naturally, you could always consider mounting the motors on the wings.
Finally, note that a key feature that allows the pitcheron or wingeron designs to work well in radio-controlled models is that there is no need to consider the aerodynamic "feel" transmitted from the control surfaces back to the control stick, or the position that the controls will go to if the pilot relaxes all forces on the control stick and just lets the control stick "float" freely.  In radio-controlled models, we are only concerned about stick-fixed stability, not stick-free stability.  Likewise we are only concerned with the control surface position that must be maintained to obtain a given result, not the force that must be applied to the control system to obtain that result. The situation is very different with full-sized airplanes, at least so long as the control system is "reversible", i.e. so long as aerodynamic forces can be transmitted from the control surfaces back to the control stick or yoke.  The same difference also applies to all-moving horizontal or vertical tails-- they are simpler to implement in radio-controlled models than in full-scale aircraft with conventional "reversible" control systems.
Footnotes:

Re "moving the wings together in the same direction would accomplish nothing in a tailless design" -- except for hang gliders and other weight-shift-controlled aircraft!  The key here, of course, is that if the CG is far enough below the rotation point of the wing (which in a hang glider is essentially the point where the pilot's "hang strap" is attached to the rigid structure of the glider), then rotating the wing (relative to the rest of the aircraft "structure", i.e. the pilot's body, or the "trike" unit in the case of a powered "trike") accomplishes a significant forward or aft translation of the CG of the whole system, relative to the wing.

Thank you! So as long as I have a fixed tail, it is possible to fly with only two movable surfaces (not counting the propeller)?

A notable example is AIM-7 Sparrow AA missile.

Thank you! The Pica looks very similar to the design I'm thinking of. Is the Sig Samurai's tail broken? I only see half of the V.

It would be nice to find a nice video of the "Pica" in action too, ideally showing a look at the internal linkages, but searching is complicated by the fact that there are apparently at least two rc slope-soarers also called "Pica" that are a much more conventional design, and also an RC flight sim program, and also a paraglider, all with the same name.  May put some more effort into it later --

@kj7rrv -- see added content re thrust line effects (now re-worked) , and control "reversibility" (last two paragraphs)

One more trivial point is that an "elevon" is mounted at the rear of the aircraft and therefore rotates in a trailing-edge-up fashion to pitch the nose up. Your proposed wing rotates in the opposite direction to pitch the nose up, because it is in front of the tail. But, "elevon" still seems an appropriate word to describe a surface that controls both pitch and roll. To be really picky/obtuse, I guess we could propose carnard+aileron = "canardon" ("caneron?") as a better name for the surface whose function you are trying to replicate by moving the wing alone?  E.g. a canard that also twists?

@quietflyer I was thinking of the movable wings being the main wings.


text:
Given Pressure Altitude and CAS, calculate Mach Number using Jeppesen E6B
Just wondering how to calculate Mach Number from CAS and PA on an E6B. For example: given FL290, CAS 280kts, Find the Mach number.
I used to use the round CR3 but I recently switched to the old-school Jeppesen E6B and found out there are far fewer functions on the front side.

@MaxPower On the CR3 there is a window that has CAS over PA which gives you a Mach number directly and you can use the Mach number and temperature to find TAS. The original question is based on a 727 that has a yaw damper failure continues to its destination and finds the TOD. Since a 727 flies faster than 200kts so compressibility became an issue thus converting directly to Mach can bypass that issue. I got 456kts TAS on the E6B where the correct answer should be around 436kts TAS

Not sure about the structure of your question.  The Mach-to-KTAS ratio is solely a function of OAT (outside air temperature) and is not a function of pressure altitude.  The speed of sound is the same at any pressure or altitude for a given OAT.  Remember that at higher speeds compressibility will make your Indicated (IAT) read higher than the actual OAT, and you may have to convert IAT to OAT).

First you have to physically “find” the Mach No Index pointer.  This is an arrow that appears in the right-side window of the wheel, the same one we use to set pressure altitude under the OAT when doing TAS calculations.  This window makes sense, because Mach Number is an OAT:TAS ratio just as pressure altitude is.  Turn the wheel so that the big Speed Index Arrow (60 on the wheel) is pointed to around 39 on the outer wheel.  You should now see Mach No Index in the Pressure Altitude window.

Place the observed OAT over the Mach No Index.

The wheel is now aligned in such a way that the outer "fixed" ring on the computer board represents KTAS and the outer-most ring of the rotating wheel represents Mach number.  We can verify we're in "the right part of town" by looking at the 10 pointer on the wheel, which represents Mach 1.0.  We know that Mach 1 will usually be in the range of 550 to 600 Kts in really cold air, to about 700 knots in really warm air.  So the 10 on the wheel ought to be between about 55 and 70.  For our examples, let say the OAT is -20c and we have set the Mach No Index under -20c.  We can see that the speed of sound (over the 10 on the wheel) is 620 Kts.

Mach to KTAS  If you know your Mach number is 0.75, then find 75 on the wheel and KTAS will be the number above it in the ring of the fixed board. You should get about 465 KTAS.

KTAS to Mach  If your KTAS is 510Kts, then find 51 on the outer ring and read Mach Number directly below it on the wheel.  You should see Mach 0.82 on the wheel.

That's it.  OAT goes over the Mach No Index in the window, and then ring is set so that Kts is over the corresponding Mach.

Because I'm feeling pedantic: Mach:TAS does change independent of OAT at extreme altitudes. Mainly because of both compositional changes in the gas mix, and non-ideal gas interactions, clouds will also have a small effect. This has no practical effect on aircraft navigation accuracy, just the uncertainty of winds aloft is orders of magnitude larger.

Yes, that’s a great point.  I mentioned compressibility in my answer, and that does have to be taken into consideration at TASes above about 200kts or so.  But you are right to point out that the ISA atmosphere is based on linear temperature and pressure lapse rates to 36,000 feet.  Above about FL360, a really accurate calculation is going to get a bit more complicated.

I was talking about the average molecular weight of the gas; and attraction/repulsion forces between the particles that are neglected in most practical formulas involving low Earth atmosphere(tropo and statosphere).


text:
Why does a sudden tailwind/headwind change IAS?
I read the following statement: “tailwind quickly changing to a headwind causes an increase in airspeed and performance.
Conversely, a headwind changing to a tailwind causes a decrease in airspeed and performance.”
I understand that’s the only situation where wind affects IAS but why does that happen?

Where did you read that statement?

@quietflyer: Since you have the rep on this site, you should be able to accept-and-edit or reject-and-edit to handle the pending edit, if you have time to also review it.

@PeterCordes -- the system let me vote to accept the edit but one more vote was needed.  Anyway that apparently has happened now, so I was now able to do the tag edit--

@quietflyer: Oh right, if you vote on an edit before deciding on making an edit yourself, then the accept-and-edit and reject-and-edit options are gone, no more chance to unilaterally deal with it if you think of something after reviewing.  I've had that happen many times :/

The word performance is a bit ambiguous here.

Why that happens is due to the aircraft's inertia.
Consider a fairly simple case: an aircraft is flying due north, straight & level, with a constant 100 knots of indicated airspeed (KIAS) and a 10 knot headwind.  So its velocity over the ground is 90 knots. There is some amount of thrust being produced, and some amount of drag, and those are in balance, just as lift & weight are in balance.  Everything is stable.
Now let's assume that the wind suddenly goes from 10 knots out of the north, to 20 knots out of the north in a very short span of time -- say, a couple of seconds.  The inertia of the aircraft is still 90 knots over the ground, but now there is more air blowing at it... that 90 knots of groundspeed and 20 knots of wind makes for 110 knots of indicated airspeed, at least at the moment that everything changed.
Assume that the pilot (or autopilot) holds altitude, but doesn't adjust the throttle.  Same thrust as before, but with the increase in airspeed you now have more drag, and so the forces are no longe in balance.  Over time, a minute or so perhaps, the airspeed will slowly decrease, since drag is greater than thrust, until you reach a new equilibrium state, back at 100 KIAS.  But now, 100 KIAS and the 20 knot headwind means the new groundspeed is 80 knots instead of 90.  Once everything gets into equilibrium, the greater headwind translates into reduced groundspeed.
But until things are back in equilibrium, the increased headwind hasn't fully reduced the groundspeed -- for a time, the greater headwind increased the airspeed, until things have a chance to get back into balance.  How long it takes for things to get back into balance depends on the magnitude of the forces involved... if your airplane was a small model airplane cruising at 15 or 20 knots with only enough thrust to overcome the fairly slight drag on the small airplane at that slow speed, then the 10-knot boost would be a huge increase in drag, and its groundspeed would bleed off within seconds as the forces quickly reached equilibrium.  On the other hand, if you have an A380 at 300 knots get the same gust, the delta in drag from 300 knots to 310 is going to be real, but comparatively smaller when you consider how much thrust & drag are already present... it won't react nearly as rapidly to that same change in the wind.  It will reach equilibrium at 10 knots less groundspeed than before, and the same IAS as it started with, but it won't do so as quickly as the model did.
Now apply that to the windshear scenario... an airplane is on approach to runway 36 (i.e. flying north) at 70 KIAS with zero headwind and on glide path, and it flies into a microburst.  The south side of the microburst has air flowing from the center of the downdraft outward, so south flow.  The airspeed increases from 70 knots to 90 knots within a few seconds.  Rather than wait for everything to get back into equilibrium, the pilot reduces power and raises the nose with the intention of holding glidepath and correcting the airspeed.  Now the drag is greater than before due to the higher airspeed, and the thrust is less than before, so the indicated airspeed is coming down quickly.  Let's say it's just back to the 70 knot target speed, when...
Now the airplane flies into the core of the down-draft, which removes the headwind and causes the airplane to sink.  And then, the airplane flies past that to the north side of the microburst, where the outflowing winds are a tailwind of 20 knots.  The aircraft's groundspeed was roughly 50 knots (back at the target 70 knots airspeed, with the 20 knot headwind, before it goes away), but now it suddenly has 20 knots of tailwind... before things get into equilibrium again, your 50 knots of groundspeed and 20 knots of tailwind means your IAS falls off to about 30 knots. You went from a 20 knot headwind to a 20 knot tailwind, a change of 40 knots.  The approach speed of 70 knots, which the pilot had just regained before losing the headwind, drops toward 30 knots -- the amount of power and trading altitude for airspeed required to avoid stalling the aircraft as the airspeed has 40 knots robbed from it within a few seconds, may be more power and altitude than is available.
Which is why windshear, and microbursts especially, are such serious business.
But the "why" behind it all is inertia.  The wind velocity can change quite rapidly, but the aircraft's velocity over the ground (or in 3-dimensional space, if you prefer) doesn't change as quickly, so the delta is reflected in the airspeed, until things are back in equilibrium.
If the 20 knot change in airspeed happened over 10 minutes, you'd hardly notice... the airspeed would start to drop off a little, the pilot would add a little power to restore his desired speed, and the higher power setting would balance the slow loss of speed, so that you'd see no loss of airspeed on the instruments, but rather a slow decrease in groundspeed.  But when the wind velocity changes suddenly, the airspeed will change faster than power changes can compensate, and you observe the gain or loss of airspeed, as described in the OP.

Let's suppose that our airplane possesses a speed of 100kts in respect to earth (speed measured for example via a GPS system) and it's flying into a windy air which has a speed of 20kts (measured for example by a ground station close to the airplane); this wind speed is also given in respect to earth.
Now, from an aerodynamic point of view, the speed which is important is the one of the mass of air investing the airplane. Let's therefore consider the case where the wind is blowing in front of the airplane toward it (headwind): in this case the airplane (and the onboard instrumentation giving the IAS) is seeing an airspeed of 100+20=120kts, which let's say is a perfectly good airspeed.
Let's suppose now that the headwind suddenly changes into a tailwind: the airplane is now seeing an airspeed of 100-20=80kts, which not only might be dangerously close to the stall speed but also changes all the aerodynamic forces... squared!

@RTO identified the source as flight-study.com. That's actually a copy of the FAA's Pilot’s Handbook of Aeronautical Knowledge, Chapter 12. The surrounding paragraphs give more context- first, it's in a section about low level wind shear. Let's see that context, emphasis mine:

While wind shear can occur at any altitude, low-level wind shear is especially hazardous due to the proximity of an aircraft to the ground....
Wind shear is dangerous to an aircraft. It can rapidly change the performance of the aircraft and disrupt the normal flight attitude. For example, a tailwind quickly changing to a headwind causes an increase in airspeed and performance. Conversely, a headwind changing to a tailwind causes a decrease in airspeed and performance. In either case, a pilot must be prepared to react immediately to these changes to maintain control of the aircraft.

So, it isn't that only sudden changes affect IAS. But it's especially critical when near to the ground and it happens suddenly.


text:
What does deploying landing flaps do to the (L/D)max and minimum drag speed of an airplane?
I thought that deploying flaps will increase both the drag and lift, thus increasing L/D ratio? But do they increase proportionately? And how exactly does that correlate with the increase/decrease of the minimum drag speed (VimD)? So I'm left with four different possibilities:

increase VimD and increase (L/D)max
increase VimD and reduce
(L/D)max
reduce VimD and increase (L/D)max
reduce VimD and
reduce (L/D)max

The first notch of flaps mainly increases lift, the last notch mainly increases drag.

I see. But the rate at which the drag increases is much higher than the rate of increase of lift thus decreasing (L/D)max?

How are you defining “minimum drag speed”?  The speed at which drag is minimum is zero- always.

@SpeedBird789 -- thought "flight dynamics" was an appropriate tag; to add it I had to pick another tag to delete.  Feel free to roll back/ undo if you prefer the tags the way they were. I know on ASE we seem to have reached the conclusion that its ok to have tags give information that's not included in the actual question so if you really want to constrain the question to General Aviation, feel free to undo or revise my tag edit.

Lowering flaps generally reduces L/D max and reduces L/D max speed

It really depends how you want to design your plane, but usually you want to have maximum excess power at your climbing speed$^1$.  This is where your plane is "clean" with its highest L/D ratio.  Or, one could design for max L/D at cruising speed, especially if they were blessed with excess power (jets) and wanted to save fuel.$^4$
But landing is generally slower, so you add flaps to increase lift, but L/D will be generally lower$^2$.
From your questions here, a better understanding can be had with the $Lift$ $Equation$:

Lift = Air Density × Wing Area × Coefficient × Velocity$^2$

Coefficient is a combination of airfoil type and AoA.  Dropping flap changes the airfoil type, increasing the lift coefficient.  So L/D max will be slower$^3$.  Greater deflection of flaps generally increases drag, lowering L/D but allowing a slower, steeper approach (bush pilots love this).
These are some guidelines, but further refinements can be made in discussion of a specific aircraft type.
$^1$ for best climbing performance
$^2$ dropping flaps creates more drag producing surfaces such as sharp corners.  Lift increases (for a given speed) but drag increases at a proportionally greater rate, giving a lower L/D.
$^3$ very high L/D ratios can be had with heavily cambered wings, but these are slower, making wind direction (groundspeed) more of an issue.
$^4$  longer runways and no speed limits at lower altitudes could make this a reality (not likely)

Just a nitpick: Dropping flaps increases lift very briefly, but mainly lowers angle of attack and increases the maximum possible lift coefficient. Since lift should equal weight, lift will stay constant over the long term.

@PeterKämpf I would posit that lowering flap (increasing coefficient) allows lower AoA (as measured along the fuselage) at the same airspeed or lower airspeed at a given AoA . Interesting here, that all things being equal, the drag increase slows the aircraft down when flaps are deployed, forcing us either to add power of increase glide angle to keep our safer AoA.

I thought that deploying flaps will increase both the drag and lift,
thus increasing L/D ratio

The lift coefficient increases, but the lift force does not, at least over the long run.  The lift force in steady-state horizontal flight must equal weight.  In a moderate climb or glide lift still must be almost equal to weight-- the actual formula is lift = weight * cosine (climb or glide angle).

But do they increase proportionately?

No they do not.  For the vast majority of (or perhaps all) airfoils used in actual aircraft, deploying the flaps (even slightly) decreases the maximum value of the ratio of ((lift coefficient) / (drag coefficient)), which is also the ratio of ((Lift force) / (Drag force)), which (in the case of an unpowered glider with zero thrust) is also the glide ratio.

And how exactly does that correlate with the increase/decrease of the
minimum drag speed (VimD)?

The speed for minimum Drag force (which is not the speed for minimum drag coefficient, but which is also the speed where the least Thrust is required to maintain constant altitude, or to maintain any given glide or climb angle) occurs at the speed for the maximum ratio of ((Lift force) / (Drag force)),  which is also the speed for the maximum ratio of ((lift coefficient) / (drag coefficient)).  Deploying flaps is somewhat like decreasing the weight of the aircraft-- due to the increased lift coefficient, the entire curve of L/D (or Cl/Cd) versus airspeed gets shifted to the left (toward lower airspeeds), as well as distorted in other ways (to reflect the decrease in max L/D or Cl/Cd), and thus the speed for the max L/D (or Cl/Cd) ratio, which is also the speed for minimum Drag force, occurs at a lower airspeed with flaps down than with flaps up.
The answer to your question is therefore option "4".
If you intended to ask about the minimum drag coefficient-- that would always correspond to high-speed flight, at a low angle-of-attack, with the flaps up.  Lowering the flaps would increase the drag coefficient at any given airspeed or airspeed or angle-of-attack, at least if we are talking about flight well above the max L/D speed.
Related ASE answers--
Is the maximum lift-drag ratio found at minimum drag? (see several answers) -- establishes that minimum Drag force occurs at the angle of attack (and logically also the flap setting) for max L/D
Can we show through simple geometry rather than formulae or graphs that the best glide ratio occurs at the maximum ratio of Lift to Drag? -- same conclusion, but geared toward gliding flight -- "Also note that the geometry of the triangle is such that the maximum Lift to Drag ratio also coincides with the minimum Drag value..."
What determines the best glide speed? -- similar


text:
Why do airplanes usually pitch nose-down in a stall?
Why do airplanes usually pitch nose-down in a fully-developed stall?
I've seen this seemingly-simple question discussed on other aviation forums, but there doesn't seem to be a single agreed-upon answer, at least that covers all or most cases.
Here is why I find this question confusing: When we say that airplanes are usually "nose-heavy," what we mean is that the center of lift is aft of the center of gravity. So the lift generated by the wings, in addition to counteracting weight, also produces a pitching moment, rotating the aircraft about its lateral axis, nose-down / tail up. This pitching moment is counteracted by the horizontal stabilizer (or canard wings) to maintain a level attitude.
In a stall, the wings produce significantly less lift. So my intuition would be that the nose-down moment would also be significantly reduced. And indeed, if the wings stalled before the tailplane, I would expect the downforce produced by the tailplane to be dominant, and the nose to pitch even further up. This is obviously contrary to experience.
Here are some of the answers I've seen proposed elsewhere on the internet:

"The CG is heavy so it falls down."

This doesn't seem like a very good answer. It's not just that the airplane is falling, it's that it's rotating, and all rotation happens about the CG. So the portion of the airplane that is pitching downward is the portion that is forward of the CG, and the portion that is pitching upward is the portion that is aft of the CG. What causes this pitching moment, as opposed to the airplane falling in its previous nose-up stalled attitude?

"The plane would fall in its previous attitude if it were in a vacuum, but because the engine is heavy, it has more inertia and is less affected by air resistance, so it falls faster."

Maybe? I don't know enough to know whether or not this is right.

"A stalled wing still produces some lift, and as it stalls, the center of lift moves aft. The greater arm results in a greater forward-pitching moment. So it is the 'nose-heaviness' being exaggerated in the stall."

This also seems plausible, but I don't know. Since stalls and stall recovery are so fundamental to basic flight instruction, I would expect this question to have a straight-forward answer, but maybe it just doesn't? Or maybe there's something I'm missing. Curious to hear people's thoughts.
Thanks!

I always figured it was because the stall reduces lift, so the aircraft begins to fall,  so the angle of attack increases, causing the tail lift to become less negative / more positive. Not that I know what I’m talking about; just another folk explanation to add to your list, I guess.

They pitch down because that's what they're designed to do.

The word in your question is "usually" and as @Pete Becker notes, it's because you design them to. Tricky to recover if it didn't.

@MikeY Yes, of course they're designed like that on purpose. My question isn't "What purpose does it serve to pitch nose-down during a stall?" but rather "What aerodynamic principles cause the nose to pitch down?"

It's actually interesting to ask the opposite question, "How would I design a reasonably flyable plane so the nose wouldn't pitch down upon wing stall."

Note that the center of pressure is also aft of the center of gravity. If you imagine a dart or arrow with a heavy point and some kind of fletching or vanes at the back, it will always want to travel through the air point first. Similar principle to a weather vane. Another way to think about it is that lift is not the only aerodynamic force on a plane. When there is no lift, the other forces remain.

Because otherwise : KSP Center of Lift Placement: Why planes and rockets flip

"How would I design a reasonably flyable plane so the nose wouldn't pitch down in a stall?"  See Wright Brothers early designs.

@MikeY: if the nose doesn't pitch down at stall it implies that it would pitch up at stall... I don't think it would be a good idea to let the nose go more and more up :)

The glib-but-maybe-actually-helpful version: because the Cp when the plane is "aerodynamically dead" and flying like a brick is behind the Cg.

@EthanB: so, did you get your answer or was it better on the other aviation forums? :D

@sophit As always, StackExchange delivers. In all seriousness, the quality and thoroughness of this discussion has been really nice to read and I feel pretty satisfied with the answers provided here.

You are quite close to the answer

And indeed, if the wings stalled before the tailplane, I would expect the downforce produced by the tailplane to be dominant, and the nose to pitch even further up. This is obviously contrary to experience.

Your assumption is that the tail keeps producing downforce. However, if the main wing stalls, the relative airflow changes. The plane goes down, so the relative airflow suddenly comes from below instead of straight ahead. The angle of attack on the tailplane increases from slightly negative (downforce) to positive. This obviously rotates the plane nose down.

I'm marking this as the answer because it's concise and makes a ton of sense. Honorable mention to @Robert-DiGiovanni, whose answer I believe corroborates this one and adds additional context and info. Thanks!

@EthanB and Sanchises This is not the whole story: yes, the airplane goes down i.e. the horizontal stabiliser sees a greater AoA but! the aircraft is also pitching nose-down i.e. tail-up and that actually decreases the AoA of the horizontal stabiliser. So there are actually two opposing effects in action.

@sophit Exactly, once you are properly nose down the tail moment is reduced and the rotation is arrested.

@EthanB and Sanchises: yes correct, I just wanted to highlight that the horizontal stabiliser is not the cause of the pitching down but actually the cure

@sophit Hm I find that hard to believe but I don't have the numbers on hand either. You'd think a deep stall would be impossible if the main wing pitching moment would be enough to put the nose down, though.

@Sanchises: I reasoned about the deep stall condition. The pitching moment after stall posses a negative slope i.e. it is "stable" in that the aircraft (or the wing at least) tends to remain in this stalled condition. And since the tailplane is shadowed by the wing's wake and doesn't work anymore, it becomes almost impossible to exit from this "stable" deep-stalled situation.

So the plane initially sinks without a change to AoA, and is designed to nose down in that event, but theoretically the phenomenon of a stall in itself needn't pitch a plane down. In fact can't some pilots, in some planes, perform a stall with the plane vertical and nose up - as a trick?

@GrimmTheOpiner That all depends on the elevator authority: how much can you pull the nose up with the air coming from below? I think most if not all planes can at least keep the nose at the horizon, basically "mushing" down, but that requires pulling the stick all the way back; the moment you release back pressure the nose drops as intended (I've flown only ASK21 gliders being a student pilot, but at the other end of the spectrum AF447 mushed down all the way into the ocean)

A properly designed aircraft must pitch down in a stall in order to lower the angle of attack of the wing, re-establishing the lift produced by the upper wing.
When an aircraft stalls, the bottom of the wing still produces some lift by simple action/reaction.  The center of pressure moves back towards the center of the wing.  The top of the wing is kaput because flow has separated.  So this mechanism and the loss of downwash on the tailplane drops the nose.
Huge amounts of forward CG are not required for this to work.
However, rotation of the nose to reduce wing AoA is a race against the aircraft's increasing vertical drop velocity, which changes the wings relative wind to favor higher AoA.
With a tiny paper airplane, rotation easily wins because the surface area to mass ratio is very high.  With a huge swept wing airliner, there is more of a tendency for the relative wind (from the drop) to change faster than the plane can rotate.

thus the legend of forward CG is born

However, the combination of rearward shift of Cp on the wing combined with reduced downforce from the tail causes the nose to drop.

relaxing elevator back pressure helps the tailplane do its job

Now we come to the dangers of abusing the rearward limit of CG.

if both the wing and rear tail plane stall, you may be in big trouble

Setting the rear tailplane "decalage" to a lower AoA than the wing allows the main wing to stall first.  However, if CG is too far back, the tailplane must produce upforce.
This creates the very real possibility that a sufficient change in relative wind can cause both wing and tailplane to stall.

if both airfoils stall, the center of vertical drag is all that is left to rotate the nose down, which is why "old time" aircraft have large, low aspect delta shaped horizontal stabilizers.

Stall behavior and recovery technique varies with aircraft, but abuse of CG limits is universally dangerous.

This answer addresses the cause-and-effect mixup that seems to have prompted the question. There is nothing about stalling that inherently causes a nose-down moment; the fact is that pushing the nose down is the fastest and most reliable way to recover from a stall, so designers make sure that their aircraft will naturally do that as a result of stall, so that their aircraft help the pilot to recover rather than fighting. It's possible to design an aircraft that doesn't pitch on stall, or pitches up, but it would be very dangerous, so designers (usually) just don't.

Comments are not for extended discussion; this conversation has been moved to chat.

It's by design so that the aircraft can recover from a stall. So that it is possible to recover from a stall, the CG of the aircraft is placed ahead of the center of lift. Suppose the CG was too far aft, the plane could fall flat (flat spin?), or be unrecoverable after a stall. There are aircraft with poor stall tendencies. A stall is a failure condition, but you do not want it to be unrecoverable.

Possibly worth adding a link to https://en.wikipedia.org/wiki/Stall_(fluid_dynamics)#Deep_stall

You don't have to design aircraft like this. You can design aircraft to do nearly anything in a stall. It's just that any option other than pitching down is incredibly dangerous (and you will not get a flight cert/flight clearance).

Why is CG in front of lift? "To provide the necessary balance between longitudinal stability and elevator control, the CG is usually located slightly forward of the center of lift. This loading condition causes a nose-down tendency in flight, which is desirable during flight at a high AOA and slow speeds." - it's an artifact of the design that also keeps the plane from flipping over in the first place. "Putting [the CoL] further behind increases stability. The closer to the CoG it is, the more maneuverable the craft becomes."

A stall is a condition where the wing (or canard, if one is fitted) exceeds the linear lift curve slope so the lift growth with increasing angle of attack is reduced and eventually reversed.
In a longitudinally stable design the tail operates at a lower lift per area than the wing and sees less angle of attack change than the wing because of the wing's downwash. This means the tail will still be comfortably in the linear range when the wing already leaves it.
Now what happens is very simple: The balance of lift between wing and tail is shifted so the tail will produce relatively more lift (in excessively stable designs less downforce), the center of pressure moves backwards and the unchanged location of the center of gravity will pull the tail up.
This nose-down rotation will produce a decrease of the local angle of attack on the tail, but only when the nose-down movement is underway. This is simple pitch damping and helps to limit the rate of rotation.

I think this is the most appropriate answer.

The wing of a GA airplane is designed so that the root of the wing (i.e. the part of the wing close to the fuselage) stalls before the wing's tips. These has mainly two positive effects:

when the root stalls, its turbulent wake impinges on the tailplane shaking it and warning the pilot of the incipient stall;
the outer part of the wing keeps on lifting and the ailerons keep on working.

Anyway to answer the question at least two other phenomenon should be taken into account:

in normal flight (i.e. not at stall) the airflow seen by the horizontal stabiliser differs from the one seen by the wing due to the downwash of the wing which decreases the AoA of the horizontal stabiliser. That means that the horizontal stabiliser has to be set to a higher AoA than if it were isolated;
we are all well aware of lift and drag coefficients (at the end of this answer I posted the plots of $C_l$ and $C_d$ for a classical airfoil like the NACA 2412, used for example for the wing of the Cessna 172). Anyway lift and drag do not tell the whole story: to completely describe the aerodynamic characteristics, a third plot is needed, which is as important as the other two but which is normally ignored: the plot of the pitching moment. For the same NACA 2412 it looks something like that (underlined in green):

What does this plot tell us? The pitching moment is normally negative i.e. nose-down. And that's why the horizontal stabiliser has to normally produce a downward lift i.e. a nose-up moment. But what happen at stall? As soon as the stall region is reached (at some 16°), the wing's pitching moment becomes suddenly almost five time more negative than just before the stall. This drop in $C_m$ sums up to a sudden drop in $C_l$ and rise in $C_d$.

Now that we have all the needed ingredients, understanding what happen approaching stall is quite easy. So, the root of the wing stalls and basically two main things happen:

a) pitching-down moment of the wing's root increases as explained in 4.;
b) the stalled wake from the wing impinges on the horizontal stabiliser shaking it as in 1. and, above all, this stalled wake doesn't generate downwash anymore as defined in 3.

The consequence of a) is that the stalled wing makes the aircraft pitch down; while the consequence of b) is that the AoA of the horizontal stabiliser increases, which lowers its downward lift, which makes the tail go up i.e. the aircraft pitch down. So, the main result of a stalled (root of the) wing is a pitch nose-down tendency.
If no corrective measures are taken, other effects follow:

lift drops... and the airplane as well;
drag rises; if the stall was entered because of a too slow velocity, higher drag will slow the airplane even more down.

Point 1. is actually not that bad as it might sound since it has two positive effects:

the fact that the airplane falls down makes it gain speed again, counteracting the drag increase of 2;
the horizontal stabiliser is (should be) designed to stall well after the wing so it is still producing lift and since the aircraft is falling down, the local AoA seen by the horizontal stabiliser increases again compensating for the decrease due to the loss of downwash as explained in b).

Here under the plot of $C_l$ (underlined in blue) and $C_d$ (in red).

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"Why do airplanes usually pitch nose-down in a fully-developed stall?"
They don't.
Many GA aircraft are designed in a way that stretches incipient stalls out way longer or that induce incipient stalls much earlier as both options increase the pilots reaction window. Some GA aircraft are designed to induce pitch down during stalls to  makes stall recovery much, much easier, but many will only drop the nose to level rather then actually pitch down. GA aircraft that are not designed to stall gracefully tend not to sell very well as they build a reputation for killing their pilots. Commercial aircraft (usually) have stall detection and warning systems so that the situations that could lead to stalls can be avoided. Actually stalling commercial aircraft outside a sim is not something that is done lightly, and as we have seen, the sim behavior often differed from real-world behavior with tragic consequence. Military and specialist aircraft have no rules.

Well, among the plethora of answers (mine included) this is actually the more logical: they do not full stall in the first place :)

I finish my comment: they do not deep stall in the first place if they are correctly designed and unless the pilot do so on purpose.

To clarify my original question slightly, I didn't mean to suggest that the nose necessarily pitched down below the horizon, just that there was typically a pitching moment in the nose-down direction. So "dropping the nose to level" would still be an example of what I'm asking about, if the plane had previously been in a 25 degree nose up attitude to bring about the stall.

@EthanB - It is not something you should rely on. Some aircraft would be just as happy to slip or spin in that situation rather then gently point their nose at or below the horizon. I should have emphasized "usually" instead of "fully-developed".

Why do airplanes usually pitch nose-down in a fully-developed stall?

Because the flight path curves downward.  It's really that simple.  There's no inherent incompatibility between the idea that the angle-of-attack of the wing is staying at a very high value (i.e. the wing is stalled), and the idea that the nose is pitching downward.
Any discussion of any "imbalance" between the lift generated by the wing and the tail, is really only addressing pitch torque, which pertains to the rotational acceleration in the pitch axis, i.e. the rate of change of pitch rotation.  But it's obvious that a non-zero rate of change of pitch rotation can only be sustained rather briefly.  Otherwise the pitch rotation rate would quickly "ramp up" to an absurdly high value.  And note that any pitch rotation automatically creates a "damping" effect that tends to alter the angle-of-attack of any surface located ahead or behind the CG, and this tends to counteract the pitch rotation-- for example, a nose-down pitch rotation makes the angle-of-attack of the tail surfaces become less positive or more negative.
So, over the long run, it's impossible to sustain an "imbalance" in "lift", or more precisely in the pitching moment generated around the CG, between the wing and the tail.  The airplane will always find some angle-of-attack, and some pitch rotation rate, where there is no such imbalance.  And so the changes in the direction of the flight path will usually be the primary driver of changes in the aircraft's pitch attitude.

This is the only correct answer in the whole list. Stalls do not cause the nose to drop. They cause a rise in nose position, (relative to aircraft flight path vector.). It only seems like there is a connection because most of the time we experience stalls from level flight, and the much larger drop in nose attitude due to the downward shift in the aircraft flight path as it drops is much more apparent.

The CG is in front of the wing (that's why),the CL is about 25 %Mac ( behind the CG )so THIS LOCATION OF THE CG MAKE the  nose be pull down like a man tripping on side walk
.Doing that the speed increases ,the AoA will decrease and laminar airflow above the top side of the wing is acquired.Pull up and fly again.

"It is normal to nose dive" does not answer why the dive happens, which is the question.

Because the Cg is forward on the aircraft , it is stated in my answer.